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Topic: test for the presence of the four anions  (Read 5412 times)

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kivine

  • Guest
test for the presence of the four anions
« on: August 25, 2004, 10:07:57 AM »
Would it be possible to test for the presence of the four anions if they were mixed together? How? Or why not?

The Chloride Ion CL-
The Iodide Ion I-
The Carbonate Ion CO3-
Sulfate Ion SO42-

Thanks

budullewraagh

  • Guest
Re:test for the presence of the four anions
« Reply #1 on: August 25, 2004, 10:24:36 AM »
for the carbonate and sulfate tests, add barium.  hopefully they will precipitate as barium sulfate and carbonate.  

as for your halides, i am not sure, but try adding silver nitrate.  i think the nitrate anion is more negative than the chloride anion and i am quite sure it is more so than the iodide anion, so you should get silver chloride and iodide, which precipitate.

kivine

  • Guest
Re:test for the presence of the four anions
« Reply #2 on: August 25, 2004, 10:52:14 AM »
this is wat my friend said, just wanna confirm whether it's correct


do QA
take out sulphate using barium..
den filter
den take out cl- and i- using silver
den filter
left with carbonate..test with acid
den separate agcl and agi by crystallisation

what kind of acid is used to test for carbonate?


Is this correct ?
Vinegar (white or red) indicates the presence of a carbonate by bubbling and fizzing.

budullewraagh

  • Guest
Re:test for the presence of the four anions
« Reply #3 on: August 25, 2004, 11:53:00 AM »
any acid will work

Demotivator

  • Guest
Re:test for the presence of the four anions
« Reply #4 on: August 25, 2004, 12:05:47 PM »
Barium will precipitate both sulfate and carbonate unless the solution is acidic. I would first slightly acidify the solution with dil nitric acid. This will release carbonate if present as carbon dioxide gas(vinegar is acetic acid so that can liberate carbonate too). Then precipitate the sulfate with barium.
Then take out Cl and I with silver nitrate. Chloride precipitate is white while iodide is yellow. If it's a mixture of both, you might not be able to tell if Cl is really present.
However, AgCl is soluble in dilute ammonia while AgI is not. So after treatment of the precipitate (if yellow) with ammonia and filtering, reacidify the liquid with nitric to precipitate any AgCl.

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